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astraxan [27]
3 years ago
6

How do you find the x intercepts of 3x^(5/3) - 4x^(7/3)

Mathematics
2 answers:
kaheart [24]3 years ago
7 0
Set the whole expression = to 0 and solve for x.

3x^(5/3) - 4x^(7/3) = 0.  Factor out x^(5/3):     x^(5/3) [3 - 4x^(2/3)] = 0

Then either x^(5/3) = 0, or 3 - 4x^(2/3) = 0.

In the latter case, 4x^(2/3) = 3. 

To solve this:  mult. both sides by x^(-2/3).  Then we have

4x^(2/3)x^(-2/3) = 3x^(-2/3),            or 4 = 3x^(-2/3).  It'd be easier to work with this if we rewrote it as

4           3
---  = --------------------
1            x^(+2/3)

Then 

4
---  = x^(-2/3).  Then, x^(2/3) = (3/4), and x = (3/4)^(3/2).  According to my     3                      calculator, that comes out to x = 0.65 (approx.)

Check this result!  subst. 0.65 for x in the given equation.  Is the equation then true?

My method here was a bit roundabout, and longer than it should have been.  Can you think of a more elegant (and shorter) solution?
Varvara68 [4.7K]3 years ago
7 0
To the risk of sounding redundant, as Altavistard already pointed out above

\bf 3x^{\frac{5}{3}}-4x^{\frac{7}{3}}=0\implies \stackrel{common~factor}{x^{\frac{5}{3}}}(3-4x^{\frac{2}{3}})=0\\\\
-------------------------------\\\\
x^{\frac{5}{3}}=0\implies \boxed{x=0}\\\\
-------------------------------\\\\
3-4x^{\frac{2}{3}}=0\implies 3=4x^{\frac{2}{3}}\implies (3)^3=\left(4x^{\frac{2}{3}}\right)^3
\\\\\\
27=64x^2\implies \cfrac{27}{64}=x^2\implies \sqrt{\cfrac{27}{64}}=x\implies \boxed{\cfrac{3\sqrt{3}}{4\sqrt{4}}=x}
\\\\\\
0.64951905283832898507\approx x
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How long will each person take to run a total of 50 miles at the given rates?
forsale [732]
Problema Solution

Terrance and Jesse are training for a long-distance race. Terrance trains at a rate of 6 miles every half hour, and Jesse trains at a rate of 2 miles every 15 minutes.

How long will each person take to run a total of 50 miles at the given rates?

Answer provided by our tutors

d = 50 mi the distance
t1 = the time of Terrance
t2 = the time of Jesse
Terrance trains at a rate of 6 miles every half hour thus his rate is:
v1 = 6/0.5 = 12 mph
Jesse trains at a rate of 2 miles every 15 minutes thus his rate is:
v2 = 2/(15/60) = 8 mph
since speed = distance/time follows time = distance/speed:
t1 = d/v1
t1 = 50/12
t1 = 4 1/6 hr
t1 = 4 hr 10 min
t2 = d/v2
t2 = 50/8
t2 = 6 1/4 hr
t2 = 6 hr 15 min
Terrance will run 4 hours and 10 minutes.
Jesse will run 6 hours and 15 minutes.

6 0
3 years ago
Please help solve the for x. <br> I'm offering 20 points
andrew11 [14]

Answer:

1/-21

Step-by-step explanation:

Step 1: Simplify both sides of the equation.

17

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−1

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Step 2: Subtract 6x from both sides.

17

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Step 3: Add 1/12 to both sides.

−7

4

x+

−1

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+

1

12

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(

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3 years ago
What is the slop of 2x+3
seraphim [82]

Answer:

2

Step-by-step explanation:

The slope, or rate of change in any function is the number multiplied by the independent variable (in this case x).

3 0
3 years ago
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Factor and solve the equation
murzikaleks [220]
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8 0
3 years ago
<img src="https://tex.z-dn.net/?f=prove%20that%5C%20%20%5Ctextless%20%5C%20br%20%2F%5C%20%20%5Ctextgreater%20%5C%20%5Cfrac%20%7B
inysia [295]

\large \bigstar \frak{ } \large\underline{\sf{Solution-}}

Consider, LHS

\begin{gathered}\rm \: \dfrac { \tan \theta + \sec \theta - 1 } { \tan \theta - \sec \theta + 1 } \\ \end{gathered}

We know,

\begin{gathered}\boxed{\sf{  \:\rm \: {sec}^{2}x - {tan}^{2}x = 1 \: \: }} \\ \end{gathered}  \\  \\  \text{So, using this identity, we get} \\  \\ \begin{gathered}\rm \: = \:\dfrac { \tan \theta + \sec \theta - ( {sec}^{2}\theta - {tan}^{2}\theta )} { \tan \theta - \sec \theta + 1 } \\ \end{gathered}

We know,

\begin{gathered}\boxed{\sf{  \:\rm \: {x}^{2} - {y}^{2} = (x + y)(x - y) \: \: }} \\ \end{gathered}  \\

So, using this identity, we get

\begin{gathered}\rm \: = \:\dfrac { \tan \theta + \sec \theta - (sec\theta + tan\theta )(sec\theta - tan\theta )} { \tan \theta - \sec \theta + 1 } \\ \end{gathered}

can be rewritten as

\begin{gathered}\rm\:=\:\dfrac {(\sec \theta + tan\theta ) - (sec\theta + tan\theta )(sec\theta -tan\theta )} { \tan \theta - \sec \theta + 1 } \\ \end{gathered} \\  \\  \\\begin{gathered}\rm \: = \:\dfrac {(\sec \theta + tan\theta ) \: \cancel{(1 - sec\theta + tan\theta )}} { \cancel{ \tan \theta - \sec \theta + 1} } \\ \end{gathered} \\  \\  \\\begin{gathered}\rm \: = \:sec\theta + tan\theta \\\end{gathered} \\  \\  \\\begin{gathered}\rm \: = \:\dfrac{1}{cos\theta } + \dfrac{sin\theta }{cos\theta } \\ \end{gathered} \\  \\  \\\begin{gathered}\rm \: = \:\dfrac{1 + sin\theta }{cos\theta } \\ \end{gathered}

<h2>Hence,</h2>

\begin{gathered} \\ \rm\implies \:\boxed{\sf{  \:\rm \: \dfrac { \tan \theta + \sec \theta - 1 } { \tan \theta - \sec \theta + 1 } = \:\dfrac{1 + sin\theta }{cos\theta } \: \: }} \\ \\ \end{gathered}

\rule{190pt}{2pt}

5 0
2 years ago
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